The toolbox contains software routines that developers use to create a consistent computer environment.
Answer:
d) daco = new Banana;
Explanation:
Dynamically allocated variables have their memory allocated in the heap memory.We declare a dynamical variable like this:-
int *a=new int ;
It means a pointer a is created on the stack memory which hold the address of the block that hold the value of variable a in heap memory.
We already have the pointer daco. We just have to initialize with keyword new.
It will be like daco=new Banana; which matches the option d.
Answer:
Binary Code
Explanation:
All microprocessors and programmable devices understand is Binary Code. These are various combinations of 0's and 1's which when placed together in a sequence represent a set of instructions that the microprocessor can read and understand to complete complex tasks. There are various other programming languages to program these tasks in an easier to read syntax for the programmers themselves but they simply take the written code and convert it into Binary before sending it to the microprocessor.
Answer:
See explanation below
Explanation:
Previos concepts
First Come First Serve (FCFS) "is an operating system scheduling algorithm that automatically executes queued requests and processes in order of their arrival".
Shortest job next (SJN), or the shortest job first (SJF) or shortest "is a scheduling policy that selects for execution the waiting process with the smallest execution time".
Shortest remaining time (SRF) "is a scheduling method that is a preemptive version of shortest job next scheduling'".
Round robin (RR) is an algorithm where the time parts "are assigned to each process in equal portions and in circular order, handling all processes without priority"
Solution for the problem
Assuming the dataset given on the plot attached.
Part a
For this algorithm the result would be:
Job A 0-6
Job B 6-(6+3) = 6-9
Job C 9-(9+1) = 9-10
Job D 10-(10+4) = 10-14
Part b
For this algorithm the result would be:
Job A 0-6
Job C 6-(6+1) = 6-7
Job B 7-(7+3) = 7-10
Job D 10-(10+4) = 10-14
Part c
For this algorithm the result would be:
Job A 0-1 until 14
Job B 2-(2+3) = 2-5
Job C 3-(3+2) = 3-5
Job D 9-(9+5) = 9-14
Part d
For this algorithm the result would be:
Job A 0-2 , 7-9, 12-14
Job B 2-4, 9-10
Job C 4-(4+1) = 4-5
Job D 5-7, 10-12